CEE 345 - Benjamin - Winter 2012 - Homework 2 Solutions(1)

CEE 345 - Benjamin - Winter 2012 - Homework 2 Solutions(1)...

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1 CEE 345 Part 2, Assignment #2 Solutions Note: A few people have asked questions about the differences in terminology between the course textbook and my notes. In the text, h a is used generically to represent the amount of head added to a fluid by a pump; thus, h a in the text is identical to TDH in my notes and is a parameter that applies to a pump independent of the system in which it is placed. Your text also uses the symbol h p to indicate the head added by the pump. However, the symbol is only used in conjunction with the energy equation for a particular system, and indicates the head provided by the pump in that system under conditions where steady operation has been achieved, i.e., at the operating point, where the energy equation is satisfied. 1. From the slides, the maximum suction lift that is allowable without risk of cavitation is: , ,R NPSH vap atm abs sl max L p p zh JJ ± ± ± ¦ Table B.1 in your text indicates that, at 100°F, the vapor pressure of water is 0.9493 psia and J = 62.00 lb/ft 3 . Also, Figure 12.12 indicates that at a flow rate of 200 gpm, NPSH R is approximately 12 ft. Thus, assuming p atm = 14.7 psia, we find: ²³ ² ³² ³ 22 2 2 , 33 14.7 lb/in 144 in /ft 0.9493 lb/in 144 in /ft 6 ft 12 ft 62.00 lb/ft 62.00 lb/ft =13.9 ft sl max z ± ± ± Since the problem states that the pump is located 12 ft above the water surface, the surface lift is less than the maximum allowable value, so no cavitation should occur. 2. Because the flow is moving between two open reservoirs, the pressure and velocity terms are all zero in the energy equation written between the surfaces of those reservoirs. The friction factor and pipe length and diameter are given, so the energy equation can be written as 2 21 2 pump lV zz h f Dg ´ ± 2 600 ft 50 ft 0.02 20 . 3 3 f t 2 * 3 2 . 2 f t / s pump V hz z f §· ± ´ ´ ¨¸ ©¹ Writing V as Q /( S D 2 /4), this expression becomes: >@ ² ³ 2 2 2 2 2 5 2 0.33 ft / 4 600 ft s 50 ft 0.02 50 ft 73.4 0.33 ft ft 2*32.2 ft/s pump Q hQ S ´ ´ Because the equation for the pump curve is given with head in feet and flow rate in gpm, it is convenient to change the coefficient on the Q 2 term in the above equation so that it also is in those units. The conversion is as follows:
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2 2 2 22 3 2 44 55 2 2 ss 1 f t 1 m i n f t - m i n f t 73.4 73.4 3.64x10 3.64x10 ft ft 7.48 gal 60 s gal gpm ±± §· § · ¨¸ ¨ ¸ ©¹ © ¹ 42 2 ft 50 ft 3.64x10 gpm pump hQ ± ² The equation of the pump curve, with TDH in ft and Q in gpm is: 2 ft TDH 180 ft 6.10x10 gpm Q ± ± The operating point occurs at the flow rate where h pump = TDH, so ft ft 50 ft 3.64x10 180 ft 6.10x10 gpm gpm QQ ² ± 365 gpm Q 3. The three data points characterizing the pump curve can be plotted as shown in the figure below, and a smooth curve can be drawn through them. We are told that when the static head is 15 m, the flow rate is 150 L/s, and the pump curve indicates that when the flow rate is 150 L/s, the pump adds 15.7 m of head to the water. Therefore, the headloss through the pipe under these conditions (equal to the sum of the static head and the head added by the pump) is 30.7 m. Because we are told that the system curve is
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CEE 345 - Benjamin - Winter 2012 - Homework 2 Solutions(1)...

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